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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Chemistry
Chapter
Redox Reactions

JEE Advanced 2015 · Paper 2 · Chemistry · Redox Reactions

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an...

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7oC{5.7}^{o}C was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0kJmol−1)( - 57.0kJmo{l}^{- 1} ) , this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka=2.0×10−5)( {K}_{a} = 2.0 \times {10}^{- 5} ) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6oC{5.6}^{o}C was measured. (Consider heat capacity of all solutions as 4.2Jg−1K−14.2J{g}^{- 1}{K}^{- 1} and density of all solutions as 1.0gmL−11.0gm{L}^{- 1} )
Enthalpy of dissociation (inkJmol−1)( inkJmo{l}^{- 1} ) of acetic acid obtained from the Expt. 2 is

Options

  1. A1.01.0
  2. B10.010.0
  3. C24.524.5
  4. D51.451.4

Correct answer

A. 1.01.0

Step-by-step solution

Let the heat capacity of insulated beaker be C.
Mass of aqueous content in Expt. 1
=(100+100)×1=200g= \left(100 + 100\right) \times 1 = 200g
⇒Totalheatcapacity=(C+200×4.2)J/K⇒ Totalheatcapacity = \left(C + 200 \times 4.2\right)J / K
Moles of acid, base neutralized in Expt. 1 =1×0.1=0.1= 1 \times 0.1 = 0.1
⇒⇒ Heat absorbed in Expt. 1 =0.1×57=5.7kJ= 0.1 \times 57 = 5.7kJ
⇒5.7×1000=(C+200×4.2)×\DeltaT⇒ 5.7 \times 1000 = \left(C + 200 \times 4.2\right) \times \DeltaT
⇒5.7×1000=(C+200+4.2)×5.7⇒ 5.7 \times 1000 = \left(C + 200 + 4.2\right) \times 5.7
⇒(C+200×4.2)=1000⇒ \left(C + 200 \times 4.2\right) = 1000
In Second experiment
nCH3COOH=0.2,nNaOH=0.1{n}_{C{H}_{3}COOH} = 0.2, {n}_{NaOH} = 0.1
Total mass of aqueous content = 200 g
⇒⇒ Total heat capacity =(C+200×4.2)=1000= \left(C + 200 \times 4.2\right) = 1000
⇒⇒ Heat released =1000×5.6=5600J= 1000 \times 5.6 = 5600J
Overall, only 0.1 mol of CH3COOHC{H}_{3}COOH undergo neutralization.
⇒ΔHneutralizationofCH3COOH=−56000.1⇒ \Delta{H}_{neutralization}of C{H}_{3}COOH = \dfrac{- 5600}{0.1}
=−56,000J/mol=−56KJ/mol= - 56,000J / mol = - 56KJ / mol
⇒ΔHionizationofCH3COOH=57−56=1KJ/mol⇒ \Delta{H}_{ionization}of C{H}_{3}COOH = 57 - 56 = 1KJ / mol
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